Q68 / 180 · KCET 2015 · Mathematics
If \( \left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0 \), then the area of the triangle whose vertices are
\[
\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right),\left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right) \text { is }
\]
\[ \left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right),\left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right) \text { is } \]
- A\( \frac{1}{4} \) abo
- B\( \frac{1}{8} a b d \)
- C\( \frac{1}{4} \)
- Correct answerD\( \frac{1}{8} \)
Solution
Given vertices of triangle \( \mathrm{ABC} \) are
\( A\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), B\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), C\left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right) \)
Now, \( \left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \rightarrow(1) \)
Taking L.H.S. of the Eq. (1) we have
\[ \left|\begin{array}{lll} 2 a & x_{1} & y_{1} \\ 2 h & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3} \end{array}\right| \]
Now \( R_{1} \rightarrow \frac{R_{1}}{a}, R_{2} \rightarrow \frac{R_{2}}{a}, R_{3} \rightarrow \frac{R_{3}}{a} \) then \( C_{1} \rightarrow \frac{C_{1}}{2} \) we have
\( \left|\begin{array}{ccc}2 & \frac{x_{1}}{a} & \frac{y_{1}}{a} \\ 2 & \frac{x_{2}}{b} & \frac{y_{2}}{b} \\ 2 & \frac{x_{3}}{c} & \frac{y_{3}}{c}\end{array}\right| \begin{array}{ccc}1 & \frac{x_{1}}{a} & \frac{y_{1}}{a} \\ 1 & \frac{x_{2}}{b} & \frac{y_{2}}{b} \\ 1 & \frac{x_{3}}{c} & \frac{y_{3}}{c}\end{array} \mid \)
We know that area od triangle is given by
\( 1 \frac{x_{1}}{a} \frac{y_{1}}{a} \)
\( A\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), B\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), C\left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right) \)
Now, \( \left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \rightarrow(1) \)
Taking L.H.S. of the Eq. (1) we have
\[ \left|\begin{array}{lll} 2 a & x_{1} & y_{1} \\ 2 h & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3} \end{array}\right| \]
Now \( R_{1} \rightarrow \frac{R_{1}}{a}, R_{2} \rightarrow \frac{R_{2}}{a}, R_{3} \rightarrow \frac{R_{3}}{a} \) then \( C_{1} \rightarrow \frac{C_{1}}{2} \) we have
\( \left|\begin{array}{ccc}2 & \frac{x_{1}}{a} & \frac{y_{1}}{a} \\ 2 & \frac{x_{2}}{b} & \frac{y_{2}}{b} \\ 2 & \frac{x_{3}}{c} & \frac{y_{3}}{c}\end{array}\right| \begin{array}{ccc}1 & \frac{x_{1}}{a} & \frac{y_{1}}{a} \\ 1 & \frac{x_{2}}{b} & \frac{y_{2}}{b} \\ 1 & \frac{x_{3}}{c} & \frac{y_{3}}{c}\end{array} \mid \)
We know that area od triangle is given by
\( 1 \frac{x_{1}}{a} \frac{y_{1}}{a} \)
Hence, the correct answer is option: 'D'.